What is area y=3cos(3x),y=3-3cos(3x),0<= x<= pi/3 ?

The answer to area y=3cos(3x),y=3-3cos(3x),0<= x<= pi/3 is -(2pi)/3+pi+2sqrt(3)

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area y=3cos(3x),y=3-3cos(3x),0<= x<= pi/3

area

y = 3 cos (3 x), y = 3 - 3 cos (3 x), 0 \leq x \leq \frac{π}{3}

- \frac{2 π}{3} + π + 23

Decimal

4.51129 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{3}} ∣ 3 cos (3 x) - (3 - 3 cos (3 x)) ∣ d x

Solve

\int_{0}^{\frac{π}{3}} ∣ 3 cos (3 x) - (3 - 3 cos (3 x)) ∣ d x : - \frac{2 π}{3} + π + 23

The area is:

= - \frac{2 π}{3} + π + 23

\frac{d}{d x} (e^{x^{3} + 4 x})

\int \frac{x ^{2} - 361}{x} d x

\int_{0}^{2} 36 x + 145 d x

t^{2} + 6 t - 17

\frac{\partial}{\partial x} (- \frac{x}{( x + y ) ^{2}})

What is area y=3cos(3x),y=3-3cos(3x),0<= x<= pi/3 ?
The answer to area y=3cos(3x),y=3-3cos(3x),0<= x<= pi/3 is -(2pi)/3+pi+2sqrt(3)